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Cipher Block Chaining - How do you change the plaintext of all blocks?

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Cipher Block Chaining - How do you change the plaintext of all blocks?

A simple block cipher based on the SHA-256 hash functionWhy, or when, to use an Initialization Vector?Precisely how does CBC mode use the initialization vector?Understanding generation of IV in CBC encryptionHow can you change this message which is in block cipher AES-256 in CBC?Cipher Feedback Mode - sizes of the blocks and shift registersCan blocks in AES-CBC ciphertext be manipulated in such a way that plaintext becomes shuffled?Cipher Block Chaining Ciphertext AlterationHow to change the known ciphertext so the known plaintext is changed in AES CBC mode?Signal Protocol Header key encryption initialization vector

1

$begingroup$

I've just read about CBC encryption and decryption and was wondering how you can change every plaintext block?

Let's say you have an initialization vector, 3 ciphertext blocks, know each block-cipher-decryption and you know each plaintext block too. Then you have some string that you want to change these given plaintexts to. Edit: Assume there is a padding oracle available as well.

If I understood correctly, it will be quite easy to change the very first plaintext block: You take each byte of the initialization vector XORed with each byte of the first plaintext block XORed with the character of the string you want to change these plaintexts to. Putting all these new bytes together, this is now the new initialization vector and XORing this new initialization vector with the first ciphertext block, we will get another plaintext; that one we wanted.

The change of the iv has no effect on the plaintext of the next ciphertext block. But now in order to change the next, namely the second plaintext, we would have to do changes on the first ciphertext block which will sadly destroy the first plaintext block... And the same will happen if we move to the next ciphertext block and try to change its plaintext. So, in the end, there will just be junk plaintext or nothing at all, bad format. But I don't see any other way where you could change all plaintext blocks correctly? : /

I've mainly oriented myself on this nice diagram here from Wikipedia:

encryption aes cbc attack

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edited 4 hours ago

kelalaka

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asked 8 hours ago

tenepolistenepolis

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1

$begingroup$

I've just read about CBC encryption and decryption and was wondering how you can change every plaintext block?

Let's say you have an initialization vector, 3 ciphertext blocks, know each block-cipher-decryption and you know each plaintext block too. Then you have some string that you want to change these given plaintexts to. Edit: Assume there is a padding oracle available as well.

If I understood correctly, it will be quite easy to change the very first plaintext block: You take each byte of the initialization vector XORed with each byte of the first plaintext block XORed with the character of the string you want to change these plaintexts to. Putting all these new bytes together, this is now the new initialization vector and XORing this new initialization vector with the first ciphertext block, we will get another plaintext; that one we wanted.

The change of the iv has no effect on the plaintext of the next ciphertext block. But now in order to change the next, namely the second plaintext, we would have to do changes on the first ciphertext block which will sadly destroy the first plaintext block... And the same will happen if we move to the next ciphertext block and try to change its plaintext. So, in the end, there will just be junk plaintext or nothing at all, bad format. But I don't see any other way where you could change all plaintext blocks correctly? : /

I've mainly oriented myself on this nice diagram here from Wikipedia:

encryption aes cbc attack

share|improve this question

edited 4 hours ago

kelalaka

9,22232352

asked 8 hours ago

tenepolistenepolis

1063

New contributor

tenepolis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I've just read about CBC encryption and decryption and was wondering how you can change every plaintext block?

Let's say you have an initialization vector, 3 ciphertext blocks, know each block-cipher-decryption and you know each plaintext block too. Then you have some string that you want to change these given plaintexts to. Edit: Assume there is a padding oracle available as well.

If I understood correctly, it will be quite easy to change the very first plaintext block: You take each byte of the initialization vector XORed with each byte of the first plaintext block XORed with the character of the string you want to change these plaintexts to. Putting all these new bytes together, this is now the new initialization vector and XORing this new initialization vector with the first ciphertext block, we will get another plaintext; that one we wanted.

The change of the iv has no effect on the plaintext of the next ciphertext block. But now in order to change the next, namely the second plaintext, we would have to do changes on the first ciphertext block which will sadly destroy the first plaintext block... And the same will happen if we move to the next ciphertext block and try to change its plaintext. So, in the end, there will just be junk plaintext or nothing at all, bad format. But I don't see any other way where you could change all plaintext blocks correctly? : /

I've mainly oriented myself on this nice diagram here from Wikipedia:

encryption aes cbc attack

share|improve this question

edited 4 hours ago

kelalaka

9,22232352

asked 8 hours ago

tenepolistenepolis

1063

New contributor

tenepolis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited 4 hours ago

kelalaka

9,22232352

asked 8 hours ago

tenepolistenepolis

1063

New contributor

tenepolis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 4 hours ago

kelalaka

9,22232352

asked 8 hours ago

tenepolistenepolis

1063

New contributor

tenepolis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 4 hours ago

kelalaka

9,22232352

edited 4 hours ago

kelalaka

9,22232352

asked 8 hours ago

tenepolistenepolis

1063

New contributor

tenepolis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

tenepolistenepolis

1063

1 Answer
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If this is all you have, then I think you can't.
You can easily get 1st and 3rd block.
Change the 2nd ciphertext to get the desired result of 3rd block and change IV to get the desired result for first.
But changing a cipher text block means you no longer have the results of the block cipher decryption.

If you have however limited access to a decryption oracle then you will just need to make trial decryption of two blocks. The ciphertext of block 2 which makes the 3rd block what you want. And then given it, you choose the ciphertext for block 1 which you send to decryption in order to calculate required IV.

share|improve this answer

answered 7 hours ago

Meir MaorMeir Maor

5,45711030

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4

$begingroup$

If this is all you have, then I think you can't.
You can easily get 1st and 3rd block.
Change the 2nd ciphertext to get the desired result of 3rd block and change IV to get the desired result for first.
But changing a cipher text block means you no longer have the results of the block cipher decryption.

If you have however limited access to a decryption oracle then you will just need to make trial decryption of two blocks. The ciphertext of block 2 which makes the 3rd block what you want. And then given it, you choose the ciphertext for block 1 which you send to decryption in order to calculate required IV.

share|improve this answer

answered 7 hours ago

Meir MaorMeir Maor

5,45711030

$endgroup$

add a comment | 

4

$begingroup$

If this is all you have, then I think you can't.
You can easily get 1st and 3rd block.
Change the 2nd ciphertext to get the desired result of 3rd block and change IV to get the desired result for first.
But changing a cipher text block means you no longer have the results of the block cipher decryption.

If you have however limited access to a decryption oracle then you will just need to make trial decryption of two blocks. The ciphertext of block 2 which makes the 3rd block what you want. And then given it, you choose the ciphertext for block 1 which you send to decryption in order to calculate required IV.

share|improve this answer

answered 7 hours ago

Meir MaorMeir Maor

5,45711030

$endgroup$

add a comment | 

$begingroup$

If this is all you have, then I think you can't.
You can easily get 1st and 3rd block.
Change the 2nd ciphertext to get the desired result of 3rd block and change IV to get the desired result for first.
But changing a cipher text block means you no longer have the results of the block cipher decryption.

If you have however limited access to a decryption oracle then you will just need to make trial decryption of two blocks. The ciphertext of block 2 which makes the 3rd block what you want. And then given it, you choose the ciphertext for block 1 which you send to decryption in order to calculate required IV.

share|improve this answer

answered 7 hours ago

Meir MaorMeir Maor

5,45711030

$endgroup$

share|improve this answer

answered 7 hours ago

Meir MaorMeir Maor

5,45711030

answered 7 hours ago

Meir MaorMeir Maor

5,45711030

answered 7 hours ago

Meir MaorMeir Maor

5,45711030